The expression is:

$$\sum_{k=1}^{n} 1 - \Bigg(\sum_{k=1}^{n} 2^{-k/2}\Bigg)^2 $$

I know that $$\sum_{k=1}^{n} 1 = 1n $$

I don't really know how to proceed with the latter part.

  • 1
    $\begingroup$ You have a geometric series $\endgroup$ – Jacky Chong Oct 19 '16 at 20:12

$$\sum_{k=1}^{n} 2^{-k/2}=\sum_{k=1}^{n} \left(\frac{1}{\sqrt{2}}\right)^k,$$ which is a geometric series (see Eq. 8). Can you take it from here? If not, please do let me know.

  • $\begingroup$ Would this change if the sum went to infinity instead of n? $\endgroup$ – Ryan Smith Oct 19 '16 at 21:31
  • $\begingroup$ @RyanSmith Yes, see Eq. 9. $\endgroup$ – Bobson Dugnutt Oct 19 '16 at 23:45
  • $\begingroup$ @Lovesovs i'm not entirely sure which one it should be actually. I asked another question with the full problem if you're interested math.stackexchange.com/questions/1976443/… $\endgroup$ – Ryan Smith Oct 19 '16 at 23:48
  • $\begingroup$ @RyanSmith Please don't just repeat an already existing question. From what you write ($k=1,2,3,\dots$), it seems $k$ runs from $1$ to $\infty$, but this also means that your variance is infinite. $\endgroup$ – Bobson Dugnutt Oct 20 '16 at 0:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.